The work done in taking an ideal gas through one cycle of operation as shown in the indicator diagram:

10^{- 5} J

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10^{- 3} J

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10^{- 2} J

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10 J

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Solution

The correct option is D $10 J$
Work done $=$ Area enclosed
$= (Δ P) (Δ V) = (4 - 2) (6 - 1) = 2 \times 5 = 10 J$

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The work done in taking an ideal gas through one cycle of operation as shown in the indicator diagram:

The correct option is D 10 JWork done = Area enclosed =(ΔP)(ΔV)=(4−2)(6−1)=2×5=10 J

The correct option is D $10 J$
Work done $=$ Area enclosed
$= (Δ P) (Δ V) = (4 - 2) (6 - 1) = 2 \times 5 = 10 J$